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  • Java - Using Linear Coordinates to Check Against AI [closed]

    - by Oliver Jones
    I'm working on some artificial intelligence, and I want my AI not to run into given coordinates as these are references of a wall/boundary. To begin with, every time my AI hits a wall, it makes a reference to that position (x,y). When it hits the same wall three times, it uses linear check points to 'imagine' there is a wall going through these coordinates. I want to now prevent my AI from going into that wall again. To detect if my coordinates make a straight line, i use: private boolean collinear(double x1, double y1, double x2, double y2, double x3, double y3) { return (y1 - y2) * (x1 - x3) == (y1 - y3) * (x1 - x2); } This returns true is the given points are linear to one another. So my problems are: How do I determine whether my robot is approaching the wall from its current trajectory? Instead of Java 'imagining' theres a line from 1, to 3. But to 'imagine' a line all the way through these linear coordinantes, until infinity (or close). I have a feeling this is going to require some confusing trigonometry? (REPOST: http://stackoverflow.com/questions/13542592/java-using-linear-coordinates-to-check-against-ai)

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  • Efficient algorithm to generate all solutions of a linear diophantine equation with ai=1

    - by Ben
    I am trying to generate all the solutions for the following equations for a given H. With H=4 : 1) ALL solutions for x_1 + x_2 + x_3 + x_4 =4 2) ALL solutions for x_1 + x_2 + x_3 = 4 3) ALL solutions for x_1 + x_2 = 4 4) ALL solutions for x_1 =4 For my problem, there are always 4 equations to solve (independently from the others). There are a total of 2^(H-1) solutions. For the previous one, here are the solutions : 1) 1 1 1 1 2) 1 1 2 and 1 2 1 and 2 1 1 3) 1 3 and 3 1 and 2 2 4) 4 Here is an R algorithm which solve the problem. library(gtools) H<-4 solutions<-NULL for(i in seq(H)) { res<-permutations(H-i+1,i,repeats.allowed=T) resum<-apply(res,1,sum) id<-which(resum==H) print(paste("solutions with ",i," variables",sep="")) print(res[id,]) } However, this algorithm makes more calculations than needed. I am sure it is possible to go faster. By that, I mean not generating the permutations for which the sums is H Any idea of a better algorithm for a given H ?

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  • Recommended library for linear programming in .Net?

    - by tbone
    Can anyone recommend a library - free, or commercial but affordable ( There are some listed here: http://en.wikipedia.org/wiki/Linear_programming#Solvers_and_scripting_.28programming.29_languages I am just starting out with LP and hope someone can recommend something. I am trying to basically minimize pricing for cell phone subscription services.

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  • Optimal two variable linear regression SQL statement (censoring outliers)

    - by Dave Jarvis
    Problem Am looking to apply the y = mx + b equation (where m is SLOPE, b is INTERCEPT) to a data set, which is retrieved as shown in the SQL code. The values from the (MySQL) query are: SLOPE = 0.0276653965651912 INTERCEPT = -57.2338357550468 SQL Code SELECT ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE, ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) - (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT FROM (SELECT D.AMOUNT, Y.YEAR FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D WHERE -- For a specific city ... -- C.ID = 8590 AND -- Find all the stations within a 15 unit radius ... -- SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) <15 AND -- Gather all known years for that station ... -- S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND -- The data before 1900 is shaky; insufficient after 2009. -- Y.YEAR BETWEEN 1900 AND 2009 AND -- Filtered by all known months ... -- M.YEAR_REF_ID = Y.ID AND -- Whittled down by category ... -- M.CATEGORY_ID = '001' AND -- Into the valid daily climate data. -- M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR ORDER BY Y.YEAR ) t Data The data is visualized here (with five outliers highlighted): Questions How do I return the y value against all rows without repeating the same query to collect and collate the data? That is, how do I "reuse" the list of t values? How would you change the query to eliminate outliers (at an 85% confidence interval)? The following results (to calculate the start and end points of the line) appear incorrect. Why are the results off by ~10 degrees (e.g., outliers skewing the data)? (1900 * 0.0276653965651912) + (-57.2338357550468) = -4.66958228 (2009 * 0.0276653965651912) + (-57.2338357550468) = -1.65405406 I would have expected the 1900 result to be around 10 (not -4.67) and the 2009 result to be around 11.50 (not -1.65). Thank you!

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  • Optimal two variable linear regression SQL statement

    - by Dave Jarvis
    Problem Am looking to apply the y = mx + b equation (where m is SLOPE, b is INTERCEPT) to a data set, which is retrieved as shown in the SQL code. The values from the (MySQL) query are: SLOPE = 0.0276653965651912 INTERCEPT = -57.2338357550468 SQL Code SELECT ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE, ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) - (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT FROM (SELECT D.AMOUNT, Y.YEAR FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D WHERE -- For a specific city ... -- C.ID = 8590 AND -- Find all the stations within a 5 unit radius ... -- SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) <15 AND -- Gather all known years for that station ... -- S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND -- The data before 1900 is shaky; and insufficient after 2009. -- Y.YEAR BETWEEN 1900 AND 2009 AND -- Filtered by all known months ... -- M.YEAR_REF_ID = Y.ID AND -- Whittled down by category ... -- M.CATEGORY_ID = '001' AND -- Into the valid daily climate data. -- M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR ORDER BY Y.YEAR ) t Data The data is visualized here: Questions How do I return the y value against all rows without repeating the same query to collect and collate the data? That is, how do I "reuse" the list of t values? How would you change the query to eliminate outliers (at an 85% confidence interval)? The following results (to calculate the start and end points of the line) appear incorrect. Why are the results off by ~10 degrees (e.g., outliers skewing the data)? (1900 * 0.0276653965651912) + (-57.2338357550468) = -4.66958228 (2009 * 0.0276653965651912) + (-57.2338357550468) = -1.65405406 I would have expected the 1900 result to be around 10 (not -4.67) and the 2009 result to be around 11.50 (not -1.65). Thank you!

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  • Optimal two variable linear regression calculation

    - by Dave Jarvis
    Problem Am looking to apply the y = mx + b equation (where m is SLOPE, b is INTERCEPT) to a data set, which is retrieved as shown in the SQL code. The values from the (MySQL) query are: SLOPE = 0.0276653965651912 INTERCEPT = -57.2338357550468 SQL Code SELECT ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE, ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) - (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT, FROM (SELECT D.AMOUNT, Y.YEAR FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D WHERE -- For a specific city ... -- C.ID = 8590 AND -- Find all the stations within a 15 unit radius ... -- SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) < 15 AND -- Gather all known years for that station ... -- S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND -- The data before 1900 is shaky; insufficient after 2009. -- Y.YEAR BETWEEN 1900 AND 2009 AND -- Filtered by all known months ... -- M.YEAR_REF_ID = Y.ID AND -- Whittled down by category ... -- M.CATEGORY_ID = '001' AND -- Into the valid daily climate data. -- M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR ORDER BY Y.YEAR ) t Data The data is visualized here: Question The following results (to calculate the start and end points of the line) appear incorrect. Why are the results off by ~10 degrees (e.g., outliers skewing the data)? (1900 * 0.0276653965651912) + (-57.2338357550468) = -4.66958228 (2009 * 0.0276653965651912) + (-57.2338357550468) = -1.65405406 I would have expected the 1900 result to be around 10 (not -4.67) and the 2009 result to be around 11.50 (not -1.65). Related Sites Least absolute deviations Robust regression Thank you!

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  • Sparse linear program solver

    - by Jacob
    This great SO answer points to a good sparse solver, but I've got constraints on x (for Ax = b) such that each element in x is >=0 an <=N. The first thing which comes to mind is an LP solver for large sparse matrices. Any ideas/recommendations?

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  • O'Reilly book clarification on 2d linear system

    - by Eric
    The Oreilly book "Learning openCV" states at page 356 : Quote Before we get totally lost, let’s consider a particular realistic situation of taking measurements on a car driving in a parking lot. We might imagine that the state of the car could be summarized by two position variables, x and y, and two velocities, vx and vy. These four variables would be the elements of the state vector xk. Th is suggests that the correct form for F is: x = [ x; y; vx; vy; ]k F = [ 1, 0, dt, 0; 0, 1, 0, dt; 0, 0, 1, 0; 0, 0, 0, 1; ] It seems natural to put 'dt' just there in the F matrix but I just don't get why. What if I have a n states system, how would I spray some "dt" in the F matrix?

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  • How to set position for a linear-gradient background in css3

    - by Virender Sehwag
    I am trying to set the position (that is, margin or padding from top) of body tag's linear background with image. My code is background-image: linear-gradient(to bottom, rgba(255, 255, 255, 0), rgba(255, 255, 255, 0), rgba(0, 0, 0, 0.9), rgb(0, 0, 0)), url("g2hd.jpg"); background-repeat: repeat, no-repeat; background-attachment: fixed; background-position: 0% 30px, center center; but 30px is not working but it works for normal for background-image:url("g2hd.jpg"); any idea

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  • Sparse quadratic program solver

    - by Jacob
    This great SO answer points to a good sparse solver, but I've got constraints on x (for Ax = b) such that each element in x is >=0 an <=N. The first thing which comes to mind is an QP solver for large sparse matrices. Also, A is huge (around 2e6x2e6) but very sparse with <=4 elements per row. Any ideas/recommendations?

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  • Linear gradients library

    - by Lieven Cardoen
    Is there a place online where I can find like 16 linear gradients that match good with each other? I need them for a chart of mine and the ones generated (by Flex) aren't good enough. So, I'm kind off searching for a library of gradients (linear in my case).

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  • Adding multiple gradients to object in Adobe Illustrator

    - by Vass
    Hi, I have an object which is a path (a nose to be specific). Now I want both a linear gradient and a radial gradient to be added to the object. So these must be separate gradient objects I guess, and I can't find a way to add multiple separate gradients to a complete path so do I duplicate the object and then apply a new gradient to each object? And what would the layer transparency features look like? Would the 'normal' overlay of the layers work? I am afraid of multiple shadows creating double dark regions, but maybe that is as its supposed to be if you think in terms of classical art and draw shadows in terms of each light obstruction.

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  • Create a linear trend line in Excel graphs with logarithmic scale

    - by Redsoft7
    I I have an Excel scatter chart with x and y values. I set the logarithmic scale in x-axis and y-axis. When I add a linear trend line to the graph, the line is not linear but appears like a curve. How can I make a linear trend line on a logarithmic-scaled chart? Sample data: x: 18449 22829 25395 36869 101419 125498 208144 2001508 14359478 17301785 y: 269,09 273,89 239,50 239,50 175,13 176,73 151,94 135,15 131,55 121,55

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  • Increasing speed of circle over time as linear with Box2d

    - by Whispered
    Assume that there is a circle and it can be moved by using keyboard arrows.Is required that increasing speed over time like increasing car speed. For example; max speed is 25 and time to reach max speed shall be 5 sec. Over 5 sec the speed will reach to max speed. Does Box2d handle that situation?. I tried setting linear valocity but it seems to make the circle have constant speed instead of increased speed over time. Thank You! Note: I'm using Box2DWeb Javascript port of Box2D.

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  • How to solve linear recurrences involving two functions?

    - by Aditya Bahuguna
    Actually I came across a question in Dynamic Programming where we need to find the number of ways to tile a 2 X N area with tiles of given dimensions.. Here is the problem statement Now after a bit of recurrence solving I came out with these. F(n) = F(n-1) + F(n-2) + 2G(n-1), and G(n) = G(n-1) + F(n-1) I know how to solve LR model where one function is there.For large N as is the case in the above problem we can do the matrix exponentiation and achieve O(k^3log(N)) time where k is the minimum number such that for all km F(n) does not depend on F(n-k). The method of solving linear recurrence with matrix exponentiation as it is given in that blog. Now for the LR involving two functions can anyone suggest an approach feasible enough for large N.

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  • Show path of a body of where it should go after linear impulse is applied

    - by Farooq Arshed
    I am making a game with Andengine and Box2D. I have a dynamic body and I apply linear impulse on the body to move it around when the user have touched the screen. Now I want to show the path where the body will go when the user have touched. If you have played Angry Birds or Basket Ball Shoot or any other which have projectile motion with a path shown you will get my point. I want to show the white dots which are shown in those games.

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  • -webkit-linear-gradient Not working in Dreamweaver CS6

    - by Ken
    I've tried multiple times to apply the following piece of code in a CSS document: display: block; width: 500px; margin: 500px auto; padding: 15px; text-align: center; border: 4px solid blue; background: -webkit-linear-gradient(top,black,white); outline: 7px solid red; Everything appears correctly, except the gradient. I have CS6 Live View turned on, and it still refuses to appear. All I get in my box is a white background, as opposed to the black to white gradient. However, when I type the same line of code into the trial of Coda 2 I downloaded, it works perfectly. Is there anything I can do to resolve the issue?

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  • Linear Search in Python? [closed]

    - by POTUS
    def find_interval(mesh,x): '''This function finds the interval containing x according to the following rules, mesh is an ordered list with n numbers return 0 if x < mesh[0] return n if mesh[n-1] < x return k if mesh[k-1] <= x < mesh[k] return n-1 if mesh[n-2] <= x <= mesh[n-1] This function does a Linear search. 08/29/2012 ''' for n in range(len(mesh)): for k in range(len(mesh)): if x == mesh[n]: print "Found x at index:" return n elif x<mesh[n]: return 0 elif mesh[n-1]<x: return n elif mesh[n-2]<=x<=mesh[n-1]: return n-1 elif mesh[k-1]<=x<mesh[k]: return k mesh = [0, 0.1, 0.25, 0.5, 0.6, 0.75, 0.9, 1] print mesh print find_interval(mesh, -1) print find_interval(mesh, 0) print find_interval(mesh, 0.1) print find_interval(mesh, 0.8) print find_interval(mesh, 0.9) print find_interval(mesh, 1) print find_interval(mesh, 1.01) Output: [0, 0.100000000000000, 0.250000000000000, 0.500000000000000, 0.600000000000000, 0.750000000000000, 0.900000000000000, 1] 0 Found x at index: 0 2 6 -1 -1 0 I don't think the output is correct. Can anyone help me fix it? Thanks.

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  • Constructive criticsm on my linear sampling Gaussian blur

    - by Aequitas
    I've been attempting to implement a gaussian blur utilising linear sampling, I've come across a few articles presented on the web and a question posed here which dealt with the topic. I've now attempted to implement my own Gaussian function and pixel shader drawing reference from these articles. This is how I'm currently calculating my weights and offsets: int support = int(sigma * 3.0) weights.push_back(exp(-(0*0)/(2*sigma*sigma))/(sqrt(2*pi)*sigma)); total += weights.back(); offsets.push_back(0); for (int i = 1; i <= support; i++) { float w1 = exp(-(i*i)/(2*sigma*sigma))/(sqrt(2*pi)*sigma); float w2 = exp(-((i+1)*(i+1))/(2*sigma*sigma))/(sqrt(2*pi)*sigma); weights.push_back(w1 + w2); total += 2.0f * weights[i]; offsets.push_back(w1 / weights[i]); } for (int i = 0; i < support; i++) { weights[i] /= total; } Here is an example of my vertical pixel shader: vec3 acc = texture2D(tex_object, v_tex_coord.st).rgb*weights[0]; vec2 pixel_size = vec2(1.0 / tex_size.x, 1.0 / tex_size.y); for (int i = 1; i < NUM_SAMPLES; i++) { acc += texture2D(tex_object, (v_tex_coord.st+(vec2(0.0, offsets[i])*pixel_size))).rgb*weights[i]; acc += texture2D(tex_object, (v_tex_coord.st-(vec2(0.0, offsets[i])*pixel_size))).rgb*weights[i]; } gl_FragColor = vec4(acc, 1.0); Am I taking the correct route with this? Any criticism or potential tips to improving my method would be much appreciated.

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  • Bounding volume hierarchy - linked nodes (linear model)

    - by teodron
    The scenario A chain of points: (Pi)i=0,N where Pi is linked to its direct neighbours (Pi-1 and Pi+1). The goal: perform efficient collision detection between any two, non-adjacent links: (PiPi+1) vs. (PjPj+1). The question: it's highly recommended in all works treating this subject of collision detection to use a broad phase and to implement it via a bounding volume hierarchy. For a chain made out of Pi nodes, it can look like this: I imagine the big blue sphere to contain all links, the green half of them, the reds a quarter and so on (the picture is not accurate, but it's there to help understand the question). What I do not understand is: How can such a hierarchy speed up computations between segments collision pairs if one has to update it for a deformable linear object such as a chain/wire/etc. each frame? More clearly, what is the actual principle of collision detection broad phases in this particular case/ how can it work when the actual computation of bounding spheres is in itself a time consuming task and has to be done (since the geometry changes) in each frame update? I think I am missing a key point - if we look at the picture where the chain is in a spiral pose, we see that most spheres are already contained within half of others or do intersect them.. it's odd if this is the way it should work.

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  • Algorithm for querying linearly through a non-linear list of questions

    - by JoshLeaves
    For a multiplayers trivia game, I need to supply my users with a new quizz in a desired subject (Science, Maths, Litt. and such) at the start of every game. I've generated about 5K quizzes for each subject and filled my database with them. So my 'Quizzes' database looks like this: |ID |Subject |Question +-----+------------+---------------------------------- | 23 |Science | What's water? | 42 |Maths | What's 2+2? | 99 |Litt. | Who wrote "Pride and Prejudice"? | 123 |Litt. | Who wrote "On The Road"? | 146 |Maths | What's 2*2? | 599 |Science | You know what's cool? |1042 |Maths | What's the Fibonacci Sequence? |1056 |Maths | What's 42? And so on... (Much more detailed/complex but I'll keep the exemple simple) As you can see, due to technical constraints (MongoDB), my IDs are not linear but I can use them as an increasing suite. So far, my algorithm to ensure two users get a new quizz when they play together is the following: // Take the last played quizzes by P1 and P2 var q_one = player_one.getLastPlayedQuizz('Maths'); var q_two = player_two.getLastPlayedQuizz('Maths'); // If both of them never played in the subject, return first quizz in the list if ((q_one == NULL) && (q_two == NULL)) return QuizzDB.findOne({subject: 'Maths'}); // If one of them never played, play the next quizz for the other player // This quizz is found by asking for the first quizz in the desired subject where // the ID is greater than the last played quizz's ID (if the last played quizz ID // is 42, this will return 146 following the above example database) if (q_one == NULL) return QuizzDB.findOne({subject: 'Maths', ID > q_two}); if (q_two == NULL) return QuizzDB.findOne({subject: 'Maths', ID > q_one}); // And if both of them have a lastPlayedQuizz, we return the next quizz for the // player whose lastPlayedQuizz got the higher ID if (q_one > q_two) return QuizzDB.findOne({subject: 'Maths', ID > q_one}); else return QuizzDB.findOne({subject: 'Maths', ID > q_two}); Now here comes the real problem: Once I get to the end of my database (let's say, P1's last played quizz in 'Maths' is 1056, P2's is 146 and P3 is 1042), following my algorithm, P1's ID is the highest so I ask for the next question in 'Maths' where ID is superior to 1056. There is nothing, so I roll back to the beginning of my quizz list (with a random skipper to avoid having the first question always show up). P1 and P2's last played will then be 42 and they will start fresh from the beginning of the list. However, if P1 (42) plays against P3 (1042), the resulting ID will be 1056...which P1 already played two games ago. Basically, players who just "rolled back" to the beginning of the list will be brought back to the end of the list by players who still haven't rolled back. The rollback WILL happen in the end, but it'll take time and there'll be a "bottleneck" at the beginning and at the end. Thus my question: What would be the best algorith to avoid this bottleneck and ensure players don't get stuck endlessly on the same quizzes? Also bear in mind that I've got some technical constraints: I can't get a random question in a subject (ie: no "QuizzDB.findOne({subject: 'Maths'}).skip(random());"). It's cool to skip on one to twenty records, but the MongoDB documentation warns against skipping too many documents. I would like to avoid building an array of every quizz played by each player and find the next non-played in the database with a $nin. Thanks for your help

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  • Gradients and memory

    - by user146780
    I'm creating a drawing application with OpenGL. I'v created an algorithm that generates gradient textures. I then map these to my polygons and this works quite well. What I realized is how much memory this requires. Creating 1000 gradients takes about 800MB and that's way too much. Is there an alternative to textures, or a way to compress them, or another way to map gradients to polygons that doesn't use up as much memory? Thanks My polygons are concave, I use GLUTesselator, and they are multicolored and point to point

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